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2 years ago

Una frase de: ama la vida quien___________________________________

Engineering

1 answer:

TEA [102]2 years ago

5 0

Answer:

A phrase from: who loves life

Explanation:

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Whenever you are around construction sites, you should ?

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Use caution when climbing on and off equipment.

Stay away from operating machinery.

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2 years ago

Mong m.n giúp mình vs ạ

Ivenika [448]

Answer:

see vous se to pe a he ko off a nack u

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1 year ago

determine the moments of inertia of the quartercircular area about the x- and y- axes, and find the polar radius of gyration abo

ASHA 777 [7]

I can’t take a full photo of step one so sorry but this is the step 2 and step 3 with the final answer

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2 years ago

Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The cha

Elza [17]

Answer:

V = 1.5062 m/s

Explanation:

look to the photos

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2 years ago

he desk has a weight of 80 lb and a centerof gravity at G. Determine the initial acceleration of a desk when the man applies eno

-BARSIC- [3]

Answer:

$N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb$

Explanation:

Na and Nb are the vertical reactions on each of the two legs at A and at B

For the horizontal forces:

$Fcos(30)-0.5N_a-0.5N_b=0\\0.5N_a+0.5N_b= Fcos(30)\\N_a+N_b= 2Fcos(30)$

For the vertical forces:

$N_a+N_b-Fsin(30)-75=0\\N_a+N_b=Fsin(30)+80$

Therefore equating both equations:

$2Fcos(30)=Fsin(30)+80\\F(2cos(30)-sin(30))=80\\F=\frac{80}{2cos(30)-sin(30)} =64.93N$

After the desk star to slide:

sum of all vertical force = ma , therefore:

$N_a+N_b-64.93sin(30)-80=0\\N_a+N_b=64.93sin(30)+80$

sum of all horizontal force = ma

$64.93cos(30)-0.2N_a-0.2N_b=\frac{80lb}{32.2ft/s^2}a\\ 0.2(N_a+N_b)=64.93cos(30)-\frac{80lb}{32.2ft/s^2}a\\N_a+N_b=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}a}{0.2}=324.65-12.42a$

equating both equations:

$324.65-12.42a=64.93sin(30)+80\\12.42a=324.65-64.93sin(30)-80\\12.42a=212.185\\a=17.08ft/s^2$

From the moment equation:

$4N_b-80(2)-64.93(3)=\frac{-80}{32.2} (17.08)(2)\\N_b=67.48lb$

$N_a=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}(17.08)}{0.2}-67.48 = 45.04lb$

For each leg: $N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb$

7 0

3 years ago